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3 years ago

An atom had a mass of 43 and it also has has 21 electrons

Chemistry

1 answer:

Serga [27]3 years ago

8 0

scandium. if you look up the number of electrons thats the periodic number and it gives you the answer

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Use water as an example to contrast the properties of a compound with the elements from which it is composed

ahrayia [7]

Properties of a compound is completely different from their elements.

Water is composed by hydrogen and oxygen.

For example, the boiling point of oxygen is - 183 °C and hydrogen is - 253 °C, meanwhile, water has a boiling point of 100°C

Another example is when you put a burning wooden splint into oxygen, it burns more brightly. Put it in hydrogen, you may hear a "pop" sound, or even explode when large amount of hydrogen. But if u put a burning splint in water, it goes off.

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2 years ago

The units used to measure heat are?

Akimi4 [234]

Answer&Explanation:

The unit used to measure Heat is joule equal to that of Energy and is abbreviated as (J)

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3 years ago

Compared to winds on the earth the winds on Saturn are much stronger. <br><br> True or False?

Yanka [14]

This is most definitly true

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3 years ago

A chemical change is like a physical change because no new substance is formed. Agree or DIsagree and reason why

aleksandrvk [35]

I disagree, because a physical change changes the form, and a chemical change is a process where one or more substances are altered into new substances.

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3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago